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给你一个按照非递减顺序排列的整数数组 nums,和一个目标值 target。请你找出给定目标值在数组中的开始位置和结束位置。
如果数组中不存在目标值 target,返回 [-1, -1]。
你必须设计并实现时间复杂度为 O(log n) 的算法解决此问题。
示例 1:
输入:nums = [5,7,7,8,8,10], target = 8
输出:[3,4]
示例 2:
输入:nums = [5,7,7,8,8,10], target = 6
输出:[-1,-1]
示例 3:
输入:nums = [], target = 0
输出:[-1,-1]
提示:
0 <= nums.length <= 105
-109 <= nums[i] <= 109
nums 是一个非递减数组
-109 <= target <= 109
class Solution {public int[] searchRange(int[] nums, int target) {int index = binarySearch(nums,target);if(index == -1){return new int[]{-1,-1};}int left = index;int right = index;while(left-1 >= 0 && nums[left-1] == target){left--;}while(right+1 <= nums.length-1 && nums[right+1] == target){right++;}return new int[]{left,right};}public int binarySearch(int[] nums,int target){int left = 0;int right = nums.length - 1;while(left <= right){int mid = left + (right-left)/2;if(nums[mid] == target){return mid;}else if(nums[mid] < target){left = mid+1;}else{right = mid-1;}}return -1;}}